CF579 - A Raisinng bacteria-白红宇

CF579 - A Raisinng bacteria

阅读量：5013 次

发布时间：2019-06-12

本文共 1648 字，大约阅读时间需要 5 分钟。

You are a lover of bacteria. You want to raise some bacteria in a box.

Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly x bacteria in the box at some moment.

What is the minimum number of bacteria you need to put into the box across those days?

Input

The only line containing one integer x (1 ≤ x ≤ 10⁹).

Output

The only line containing one integer: the answer.

Examples

Input

Output

Input

Output

Note

For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.

For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1.

题解：水题；

参考代码：

1 #include
      
        2 #include
       
         3 #include
        
          4 #include
         
           5 #include
          
            6 #include
           
             7 #include
            
              8 #include
             
               9 #include
              
               10 #include
               
                11 #include
                
                 12 #include
                 13 using namespace std;14 typedef long long LL;15 typedef pair
                  
                    PII;16 #define PI acos(-1)17 #define EPS 1e-818 const int INF=0x3f3f3f3f;19 const LL inf=0x3f3f3f3f3f3f3f3fLL;20 const int maxn=1e5+10;21 22 int main()23 {24 int n;25 while(~scanf("%d",&n))26 {27 int cnt=0;28 for(int t=1;t<=n;t*=2)29 {30 if((t&n)!=0) cnt++,n-=t;31 } 32 printf("%d\n",cnt); 33 }34 return 0;35 }

View Code

转载于:https://www.cnblogs.com/songorz/p/9520420.html

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